Proof: By Euclid

For let $AD$ have been produced in both directions to $G$ and $H$, and let the (straight line) $BG$ have been drawn through $B$ parallel to $CA$ [Prop. 1.31], and let the (straight line) $FH$ have been drawn through $F$ parallel to $DE$ [Prop. 1.31].
Thus, $GBCA$ and $DEFH$ are each parallelograms.
And $GBCA$ is equal to $DEFH$.
For they are on the equal bases $BC$ and $EF$, and between the same parallels $BF$ and $GH$ [Prop. 1.36].
And triangle $ABC$ is half of the parallelogram $GBCA$.
For the diagonal $AB$ cuts the latter in half [Prop. 1.34].
And triangle $FED$ (is) half of parallelogram $DEFH$.
For the diagonal $DF$ cuts the latter in half.
And the halves of equal things are equal to one another. Thus, triangle $ABC$ is equal to triangle $DEF$.
Thus, triangles which are on equal bases and between the same parallels are equal to one another.
(Which is) the very thing it was required to show.
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